2nd PUC Maths Question Bank Chapter 10 Vector Algebra

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Karnataka 2nd PUC Maths Question Bank Chapter 10 Vector Algebra

2nd PUC Maths Vector Algebra One Marks Questions and Answers

Question 1.
Compute the magnitude of the following vectors

Answer:

Question 2.
Write two different vectors having same magnitude.
Answer:
Two vectors can have same magnitude, if the sum of the squares of coefficient of î ĵ and k̂ is same. Let vectors a = (2î + 3ĵ + k̂) and b = (2î + 3ĵ + k̂) are different vectors having the same magnitude.

i.e., they have same magnitude.

Question 3.
Write two different vectors having same direction.
Answer:

Question 4.
Find the value of and x so that the vectors 2î + 3ĵ and xî + yĵ are equal.
Answer:
Let a = 2î + 3ĵ and b = xî + yĵ be two given vectors.
Comparing coefficients of î and ĵ of both a and b, we have
⇒ x = 2 and y = 3.

Question 5.
Find the scalar and vector components of the vector with initial point (2,1) and terminal point (-5,7).
Answer:
Vector with initial point A(2,1) and final (terminal) point B (-5,7) can be given by
AB = (x₂ – x₁)î + (y₂ – y₁)ĵ = (-5 – 2)î + (1 – 1)î = (-7)î + 6ĵ
Hence, the required scalar components (coefficients of î and ĵ) re -7 and 6 while the vector components are 7î and 6ĵ.

2nd PUC Maths Vector Algebra Two Marks/Three Marks Questions and Answers

Question 1.
Find the sum of the vectors a = î – 2ĵ + k̂, b = -2î + 4ĵ + 5k̂ and c = î – 6ĵ – 7k̂
Answer:
Here, given a = î – 2ĵ + k̂, b = -2î + 4ĵ + 5k̂, c = î – 6ĵ – 7k̂ Sum of these vectors can be calculated by adding their î, ĵ and k̂ components.

Question 2.
Find the unit vector in the direction of the vector a = î + ĵ + 2k̂.
Answer:

Question 3.
Find the unit vector in the direction of vector \(\overrightarrow{\mathrm{PQ}}\), where P and Q are the points (1, 2, 6). Q = (4, 5, 6) respectively.
Answer:
The given points are P( 1,2,3) and Q (4,5,6).
∴ x₁ = 1, y₁ = 2, z₁ = 3 and x₂ = 4, y₂ = 5, z₂ = 6

Question 4.
For given vectors, a = 2î – ĵ + 2k̂ and b = -î + ĵ – k̂, find the unit vector in the direction of the vector a + b.
Answer:

Question 5.
Find a vector in the direction of vector 5î – ĵ + 2k̂ 2A which has magnitude 8 unit.
Answer:
Let \(\vec{a}\) = 5î – ĵ + 2k̂
Compring with X = xî – yĵ + zk̂, we get x = 5, y = -1 z = 2

Question 6.
Show that the vectors 2î – 3ĵ + 4k̂ and -4î + 6ĵ – 8k̂ are collinear.
Answer:

Vectors a and b have the same direction, therefore they are collinear.

Question 7.
Find the direction cosines of the vector î + 2ĵ + 3k̂.
Answer:

Question 8.
Find the direction cosines of the vector joining the points A (1,2,-3) and B (-1,-2, 1), directed from A to B.
Answer:
The given points are A (1,2, -3) and B (-1,-2,1).
i.e., x₁ = 1, y₁ = 2, z₁ =- 3 and x₂ = – 1, z₂ = 1

Question 9.
Show tht the vector î + ĵ + k̂ is equally inclined to the axes OX, OY and OZ.
Answer:
Let a = î + ĵ + k̂

If a makes angles a, b and r respectively with (positive) OX, OY and OZ.
Then, we have cos α = 1 /√3
(∵ Direction cosines are the cosines of the angles made by î, ĵ, k̂ components of the vector with X, Y, and Z axes)

Hence, the given vector a is equally inclined with OX, 0 Y and OZ.

Question 10.
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are î + 2ĵ – k̂ and -î + ĵ – k̂ respectively, in the ration 2 : 1
(i) internally
(ii) externally.
Answer:
The position vector of a point R divided the line segment joining two points P and Q in the ratio m:n is given by

Question 11.
Fnd the position vector of the mid-point of the vector joining points P (2,3,4) and Q(4,1,-2).
Answer:
The position vector of mid-point of the vector joining the points P(2,3,4) and Q(4,1,-2) is given by,
PVof the mid-point of (PQ) = \(\frac { 1 }{ 2 }\)(PV of P + PVof Q)

Question 12.
Show that the points A, B and C with position vectors, \(\vec{a}\) = 3î – 4ĵ – 4k̂, \(\vec{b}\) = 2î – ĵ + k̂ and \(\vec{c}\) = î – 3ĵ – 5k̂ respectively, form the vertices of a right anled triangle.
Answer:
Position vectors of points A, Band Care respectively given as

Therefore, ∆ ABC is a right anged triangle with right angle at A.

Question 13.
Find the angle between two vectors a and b with magnitudes √3 and 2 respectively, having a.b = √6.
Answer:
It is given that |a| = √3, |b| = 2 and a.b = √6
Let θ be the required angle, then

Hence, the angle between the given vectors a and b is π/4.

Question 14.
Find the angle between the vectors î – 2ĵ + 3k̂ and 3î – 3ĵ – 5k̂
Answer:
Let \(\vec{a}\) = î – 2ĵ + 3k̂ and \(\vec{b}\) = 3î – 3ĵ – 5k̂

= 1.3 + (-2).(-2) + 3.1 = 3 + 4 + 3 = 10
(Dot product of two vectors is equal to the sum of the products of their corresponding components.)
Let θ be the required angle between a and b, then

Question 15.
Find the projection of the vector î – ĵ on the vector î + ĵ .
Answer
Let a = î – ĵ, b = î + ĵ

Hence, the proection of vector \(\vec{a}\) on \(\vec{b}\) is 0.

Question 16.
Find the projection of the vector î + 3ĵ + 7k̂ A on the vector 7î – ĵ + 8k̂.
Answer:

Question 17.

Answer:

Question 18.
Evaluate the product (3\(\vec{a}\) – 5\(\vec{b}\)).(2\(\vec{a}\) + 7\(\vec{b}\)).
Answer:
We have, (3\(\vec{a}\) – 5\(\vec{b}\)).(2\(\vec{a}\) + 7\(\vec{b}\))
= (3a).(2a + 7b) – (5\(\vec{b}\)).{2\(\vec{a}\) + 7\(\vec{b}\))
= 6 (a.a) + 21(a.b) – 10 (b.a) – 35 (b.b)
= 6 |a|² + 11 (a.b) – 35 |b|² (∵ a.a = |a|² and a.b = b.a)

Question 19.
Find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\) having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.
Answer:
Both vectors have same magnitude i.e.,|a| = |b|
and scalar product of vectors, \(\vec{a}\).\(\vec{b}\) = \(\frac { 1 }{ 2 }\) (Given)
Let θ be the angle between two vectors a and b, then

Question 20.
Find |x|, if for a unit vector a, (\(\vec{x}\) – \(\vec{a}\)).(\(\vec{x}\) + \(\vec{a}\)) = 12.
Answer:
Given |a| = 1
(∵ It is a unit vector, magnitude of a unit vector is 1.)

Question 21.
If a = 2î + 2ĵ + 3k̂, b = -î + 2ĵ + k̂ and c = 3î + ĵ such that a + λb is perpendicular to c, then And the value of λ.
Answer:
The given vectors are a = 2î + 2ĵ + 3k̂, b = -î + 2ĵ + k̂ and c = 3î + ĵ
Now, (a + λb) ⊥ c (Given)
⇒ (a + λb).c = 0
(∵ scalar product of two perpendicular vectors is zero)

Hence, the required value of λ is 8.

Question 22.
Show that |a| b + |b| a is perpendicular to |a| b – |b| a for any two non-zero vectors a and b.
Answer:
Let p = |a| b + |b| a is perpendicular to |a|b – |b| a
Then p.q = (|a| b + |b|a).(|a| b – |b|a)
= |a|² (b.b) – |a| |b| (b.a) + |b| |a| (a.b) – |b|² (a.a)
= |a|² |b|² – |a| |b| (a.b) + |a| |b\ (a.b) – |b|² |a|² = 0
⇒ P ⊥ q (∵ If c.d = 0 ⇒ c is perpendicular to d)
Hence, |a| b + |b|a and |a|b – |b|a are perpendicular to each other for any two non-zero vectors a and b.

Question 23.
If a, b, c are unit vectors such that a + b + c = 0, then find the value of a.b + b.c + c.a.
Answer:
Given, |a| = |b| – |c| = 1 and a + b + c = 0
We have (a + b + c).(a + b + c) = 0
⇒ a.(a + b + c) + b .(a + b + c) + c.(a + b + c) = 0
⇒ a.a + a.b + a.c + b.a + b.b + b.c + c.a + c.b + c.c = 0

⇒ 3 + 2 (a.b + b.c + c.a) = 0
⇒ a.b + b.c + c.a = –\(\frac { -3 }{ 2 }\).

Question 24.
If either a = 0 or b = 0, then a.b = 0. But the converse need not to be true. Justify your answer with an example.
Answer:
If a = 0 = 0î + 0ĵ + 0k̂ and b is non-zero i.e., b = xî + yĵ + zk̂
∴ a.b = (0î + 0ĵ + 0k̂) (xî + yĵ + zk̂) = (0 × t) + (0 × y) + (0 × z) = 0
So, if a = 0 or b = 0, then for same a.b = 0
To prove that converseneed not to be we have to prove that for two non-zero vectors a and’ b, a.b can be zero.

Hence, the converse of the given statement need not be true.

Question 25.
If the vertices A, B, C of a triangle ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively then find ∠ABC (∠ABC si the angle between the vectors BA and BC).
Answer:
We are given the points A(1, 2, 3), B(-1, 0, 0) and C(0, 1, 2).
Also, it is given that ∠ABC is the angle between the vectors BA and BC.
Here,

Question 26.
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear.
Answer:
The given points re A(1, 2, 7), B(2, 6,3) and C(3, 10, -1).

∴ |AC| = |AB| + |BC|. Hence, the given points A, B and C are coliinear.

Question 27.
Show that the vectors 2î – ĵ + k̂, î – 3ĵ – 5k̂ and 3î – 4ĵ – 4k̂ form the vertices of a right angled triangle.
Answer:
Let 2î – ĵ + k̂, î – 3ĵ – 5k̂ and 3î – 4ĵ – 4k̂

Question 28.
Find \(|\vec{a} \times \vec{b}|\), if a = î – 7ĵ + 7k̂ and b = 3î – 2ĵ + 2k̂
Answer:
It is given that a = î – 7ĵ + 7k̂ and b = 3î – 2ĵ + 2k̂

Question 29.
Find a unit vector perpendicular to each of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\), where \(\vec{a}\) = 3î + 2ĵ + 2k̂ \(\vec{b}\) = î + 2ĵ – 2k̂.
Answer:

Question 30.
If a unit vector \(\vec{a}\), makes angle \(\frac{\pi}{3}\) with î. \(\frac{\pi}{4}\) with ĵ and an acute angle θ with k̂, then find θ and hence the components of a.
Answer:

Question 31.

Answer:

Question 32.
Find λ and u, if (2î + 6ĵ + 27k̂) × (î + λĵ + µk̂) = 0
Answer:

Question 33.
Given that a.b = 0 and a × b = 0. What can you conclude about the vectors a and b?
Answer:
Given that a.b = 0
Then, either |a| = 0 or |b| = 0 or a ⊥ b (in case a and b are non-zero) and if a × b = 0 then either |a| = 0 or |b| = 0or a||b (in case a and b are non-zero) But a and b cannot be perpendicular and parallel simulataneously.
Hence, |a| = 0 or |b| = 0.

Question 34.

Answer:

Question 35.

Answer:

Question 36.
Find the volume of the parallelopiped whose cotermius edges are 2î + ĵ + 3k̂, – î + 2ĵ + k̂ and 3î + ĵ + 2k̂.
Answer:
Let \(\vec{a}\) = 2î + ĵ + 3k̂, \(\vec{b}\) = – î + 2ĵ + k̂ and \(\vec{c}\) = 3î + ĵ + 2k̂.

Question 37.
Show that the vectors
\(\vec{a}\) = î – 2ĵ + 3k̂, \(\vec{b}\) = – 2î + 3ĵ – 4k̂ and \(\vec{c}\) = î – 3ĵ + 5k̂are coplanar.
Answer: